Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 508: 26

Answer

$$\int_{0}^{1}\frac{3x^{2}+1}{x^{3}+x^{2}+x+1}dx=\frac{5}{2}ln2-\frac{\pi}{4}$$

Work Step by Step

$$\int_{0}^{1}\frac{3x^{2}+1}{x^{3}+x^{2}+x+1}dx=\int_{0}^{1}(\frac{2}{x+1}+\frac{x-1}{x^{2}+1})dx$$ $$=\int_{0}^{1}(\frac{2}{x+1}+\frac{x}{x^{2}+1}-\frac{1}{x^{2}+1})dx$$ $$=\int_{0}^{1}(\frac{2}{x+1}-\frac{1}{x^{2}+1})dx+\frac{1}{2}\int_{0}^{1}\frac{1}{x^{2}+1}d(x^{2})$$ $$\left [2ln(x+1)-arctan\,x\right ]_{0}^{1}+\left [\frac{ln(x^{2}+1)}{2}\right ]_{0}^{1}=\frac{5}{2}ln2-\frac{\pi}{4}$$
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